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leetcode(4) median of two sorted arrays 解法：

admin 11月 30, 2020 0

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

解法：

我采用的解法是用2个变量分别指向两个数组，每次取较小的一个，然后将其指针后移动，直至找到中位数

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class :
    def findMedianSortedArrays(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: float
        """
        if(len(nums1) == 0):
            if(len(nums2)%2 != 0):
                return nums2[int(len(nums2)/2)];
            else:
                result1 = nums2[int(len(nums2)/2-1)];
                result = nums2[int(len(nums2)/2)];
                result = (result+result1)/2;
                return result;
        if (len(nums2) == 0):
            if (len(nums1) % 2 != 0):
                return nums1[int(len(nums1) / 2)];
            else:
                result1 = nums1[int(len(nums1) / 2 - 1)];
                result = nums1[int(len(nums1) / 2)];
                result = (result + result1) / 2;
                return result;
        t = len(nums1) + len(nums2);
        if (t % 2 != 0):
            pos = t / 2;
            flag = 0;
        else:
            pos1 = t / 2 - 1;
            pos2 = t / 2;
            flag = 1;
        i = 0;
        j = 0;
        k = 0;
        result = -1;
        while (i < len(nums1) or j < len(nums2)):
            if(i == len(nums1)):
                result = nums2[j];
                j+=1;k+=1;
            elif(j == len(nums2)):
                result = nums1[i];
                i += 1;
                k += 1;

            elif (nums1[i] < nums2[j]):
                result = nums1[i];
                i += 1;
                k += 1;
            elif (nums1[i] > nums2[j]):
                result = nums2[j];
                j += 1;
                k += 1;
            else:
                result = nums1[i];
                i += 1;
                k += 2;
                j += 1;
            if (flag == 0 and k > pos):
                return result;
            if (flag == 1 and k > pos1):
                if(i == len(nums1) and j == len(nums2)):
                    return result;
                if(i == len(nums1)):
                    result1 = nums2[j];
                    result = (result + result1) / 2.0;
                    return result;
                if(j == len(nums2)):
                    result1 = nums1[i];
                    result = (result + result1) / 2.0;
                    return result;
                if (nums1[i] < nums2[j]):
                    result1 = nums1[i];
                else:
                    result1 = nums2[j];
                result = (result + result1) / 2.0;
                return result;

还可以采用分治法解决该问题 参考链接：http://blog.csdn.net/hk2291976/article/details/51107778

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